\documentclass[12pt]{article}
\author{William A. Stein}
\title{Algebraic Geometry Homework\\
II.3 \#'s 6,7,8,12,14\\ II.4 \#'s 2, 4}
\font\grmn=eufm10 scaled \magstep 1
\newcommand{\gp}{p}
\newcommand{\ga}{a}
\newcommand{\so}{\mathcal{O}}
\newcommand{\id}{\mbox{\rm id}}
\newtheorem{prob}{Problem}
\newtheorem{prop}{Proposition}
\newtheorem{theorem}{Theorem}
\newcommand{\ov}{\overline{\varphi}}
\newcommand{\ol}{\overline}
\renewcommand{\phi}{\varphi}
\newcommand{\fh}{f^{\#}}
\newcommand{\proj}{\mbox{\rm Proj \hspace{.01in}}}
\newcommand{\spec}{\mbox{\rm Spec \hspace{.01in}}}
\newcommand{\proof}{\mbox{\sc Proof.\hspace{.1in}}}
\newcommand{\diagram}{$$\vspace{.75in}$$} %% space for diagrams
\begin{document}
\maketitle
\begin{prob}[3.6]
Let $X$ be an integral scheme. Show that the local ring $\so_{\xi}$ of
the generic point $\xi$ of $X$ is a field. Call it $K(X)$. Show also
that if $U=\spec A$ is any open affine subset of $X$, then $K(X)$ is
isomorphic to the quotient field of $A$.
\end{prob}
\proof
Let $U=\spec A$ be any nonempty open affine subset of $X$. Then since
the closure of a generic point of $X$ is all of $X$, every open set
must contain a generic point. Thus if $\xi$ is a generic point, then
$\xi\in U$. But $A$ is an integral domain so $(0)$ is the unique
generic point of $U$, whence $\xi=(0)$. This shows the generic point is
unique if it exists. Since $X$ is integral it is irreducible so every
open set intersects $U$. Thus every open set contains $(0)\in \spec A=U$,
so $X$ actually contains a generic point $\xi=(0)$. Furthermore,
$\so_{\xi}\cong A_{(0)}$ is the quotient field of $A$.
\begin{prob}[3.7]
Let $f:X\rightarrow Y$ be a dominant, generically finite morphism of
finite type of integral schemes. Show that there is an open dense
subset $U\subseteq Y$ such that the induced morphism
$f^{-1}(U)\rightarrow U$ is finite.
\end{prob}
\proof
Let $U=\spec B$ be an open affine subset of $Y$ which contains the
generic point of $Y$. Let $V=\spec A$ be an open affine subset of
$f^{-1}(U)$. Since $f$ is of finite type $A$ is a finitely generated
$B$-algebra. The generic point of $X$ is in $V$ since every
open set contains the generic point. Let $\phi:B\rightarrow A$ be
the homomorphism corresponding to the induced morphism of affine schemes
$f:V\rightarrow U$.
Since $f$ is dominant, we know that $\phi$ is injective.
The induced map on stalks then gives an inclusion
of function fields $K(Y)=B_{(0)}\hookrightarrow A_{(0)}=K(X)$.
Since $A$ is a finitely generated $B$-algebra, $K(X)$ is a finitely
generated field extension of $K(Y)$. If this field extension is not of
finite degree then $K(X)$ must contain an element which is
transcendental over $K(Y)$. Thus $A$ must contain an
element $t$ which is transcendental over $B$. But then infinitely
many primes of $A$ lie over $(0)$. Indeed, since $K(X)$ is
finitely generated over $K(Y)$, $K(X)$ is
a finite algebraic extension of $k(t)$ for some field $k$.
Then (since the algebraic closure of a field is
infinite), there are infinitely many irreducible polynomials
in $k[t]$. Since $K(X)$ is finite algebraic over $k(t)$
infinitely many of these must remain irreducible in $A$.
Multiplying through denominators this gives infinitely
many irreducible elements of $A$ which generate prime ideals
which lie over $(0)$. This would contradict the fact that
$f$ is generically finite. Thus $K(X)$ is finite over $K(Y)$.
Let $x_1,\ldots,x_n$ generate $A$ as a $B$-algebra. Then,
since $K(X)$ is finite over $K(Y)$ each $x_i$ satisfies
some polynomial $f_i$ with coefficients in $B$. Let $b$ be the
product of all of the leading coefficients of the polynomials
$f_i$. If $b$ is a unit in $B$ then so are all of the leading
coefficients of the $f_i$ so we can divide by them and hence assume
the $f_i$ are monic polynomials. If not, replace $B$ by the localization
$B_b$ and repeat the whole argument with $U=\spec B_b$.
In either case we may assume the $f_i$ are monic from which
we conclude that $A$ is a finitely generated integral extension
of $B$, thus $A$ is a finite module over $B$.
Now let $U=\spec B$ be an open affine subset of $Y$ which contains the
generic point. Since $f$ is of finite type we may write
$f^{-1}(U)=\cup_{i=1,\ldots,n}V_i$ where each $V_i=\spec A_i$ is
finitely generated $B$-algebra. By the work above we may shrink
$U$ so that we can assume each $A_i$ is actually a finitely
generated $B$-module. To complete the proof we need to show that
there is a distinguised open subset of $U$ (which necessarily
contains the generic point) whose inverse image under $f$ is an
open affine which is the spectrum of a finitely generated $B$-module.
Let $\phi_i:B\rightarrow A_i$ be the homomorphism which
induces $f|_{V_i}$. Since $f$ is dominant, each $\phi_i$
is an injection. Thus we may, for notational convenience,
identify $B$ with it's images in the various $A_i$. The
morphism $f$ is then induced by the inclusion map
$B\hookrightarrow A_i$.
Since $\cap_{i=1,\ldots,n}V_i$ is open we can,
for each $i$, $1\leq i \leq n-1$, find $\alpha_i\in A_i$
such that $\spec (A_i)_{\alpha_i}\subseteq \cap_{i=1,\ldots,n}V_i$.
Since $A_i$ is a finite module over $B$ there is an
integral equation
$$\alpha_i^n+b_{n-1}\alpha_i^{n-1}+\cdots+b_0=0$$
where each $b_j\in B$ and $b_0\not=0$.
Let $b=\prod_{i=1,\ldots,n-1}b_i$. Then any
prime of $A_i$ which contains $\alpha_i$ must
also contain $b_i$ and hence $b$. Therefore
$\spec (A_i)_{b}\subseteq \spec (A_i)_{\alpha}$.
We then have that
$g^{-1}(\spec B_b)
= \cup_{i=1,\ldots,n-1}\spec (A_i)_b\cup\spec(A_n)_b
= \spec(A_n)_b$.
The latter equality follows since, for $1\leq i\leq n-1$,
$\spec(A_i)_b\subseteq \cap_{i=1,\ldots,n} V_i \cap f^{-1}(U_b)
\subseteq f^{-1}(U_b)\cap V_n = \spec (A_n)_b$.
We thus see that $U_b$ is a dense open
subset of $Y$ such that the morphism $f:f^{-1}(U_b)\rightarrow U_b$
is finite (since $f^{-1}(U_b)$ is the affine scheme $\spec(A_n)_b$
which a finite $B_b$-module). This completes the proof.
\begin{prob}[3.8]
Normalization. Let $X$ be an integral scheme. For each open affine
subset $U=\spec A$, let $\tilde A$ be the integral closure of $A$
in its quotient field, and let $\tilde U=\spec \tilde A$.
Show that one can glue the schemes $\tilde U$ to obtain a normal integral
scheme $\tilde X$, called the normalization of $X$. Show also that there
is a morphism $\tilde X\rightarrow X$, having the following universal
property: for every normal integral scheme $Z$, and for every
dominant morphism $f:Z\rightarrow X$, $f$ factors uniquely through
$\tilde X$. If $X$ is of finite type over a field $k$, then the morphism
$\tilde X\rightarrow X$ is a finite morphism.
\end{prob}
\proof
We first verify the universal property for affine schemes where
it is clear what the normalization is.
\begin{prop} Suppose $X=\spec A$, $\tilde X=\spec \tilde A$
its normalization and $Z=\spec B$ is a normal integral scheme.
Then every dominant morphism $f:Z\rightarrow X$ factors uniquely
through $\tilde X$.
\end{prop}
\proof
Let $\phi:A\rightarrow B$ be the homomorphism corresponding to $f$.
Then, since $f$ is dominant, $\phi$ is injective. Indeed,
$f(Z)\subseteq V(\ker(\phi))$ so if $\ker(\phi)\not = 0$
then $f(Z)$ doesn't meet the nonempty open set
$X-V(\ker(\phi))$ (nonempty since $A$ is a domain
so $(0)$ is prime so $(0)\in X-V(\ker(\phi))$.
So there is a unique extension of $\phi$ to a
homomorphism from $\tilde A \rightarrow B$.
Indeed, define $\ol{\phi}(a/b)=\phi(a)/\phi(b)$. Then
since $\phi$ is injective this is well-defined since $b\not=0$
implies $\phi(b)\not=0$. Furthermore, if $\psi$ is another
possible extension of $\phi$ to $\tilde A$, then
$\psi(b)\psi(a/b)=\psi(a)=\phi(a)=\phi(b)\ol{\phi}(a/b)=\psi(b)\ol{\phi}(a/b)$
so cancelling shows that $\psi(a/b)=\ol{\phi}(a/b)$.
Thus there is a unique morphism $f':Z\rightarrow
\tilde X$ whose composition with the natural map $\tilde X\rightarrow X$
equals $f$. (The natural map is induced by the inclusion
$A\hookrightarrow \tilde A$.)
Now we prove the universal property holds when $Z$ is an arbitrary
normal integral scheme but $X$ is still affine.
\begin{prop} Let $X=\spec A$, $\tilde X=\spec \tilde A$ its
normalization and $Z$ be any normal integral scheme. Then
every dominant morphism $f:Z\rightarrow X$ factors uniquely
through $\tilde X$.
\end{prop}
\proof
Let $U_i$ be a cover of $Z$ by open affines. If $U=U_i$ is any
$U_i$ then $U$ is a normal
integral affine scheme and $f|_U$ is a dominant morphism. Indeed,
$U$ is dense in $Z$ since $Z$ is irreducible (Proposition 3.1).
Thus $f^{-1}(\ol{f(U)})\supseteq \ol{U}=Z$ so
$f^{-1}(\ol{f(U)})=Z$ so $\ol{f(U)}\supseteq f(Z)$ so
$\ol{f(U)}=\ol{f(Z)}=X$.
We can thus apply the above proposition to find a unique morphism
$g_i:U_i\rightarrow \tilde{X}$ such that $\psi\circ g_i=f|_U$ where
$\psi:\tilde{X}\rightarrow X$. By uniqueness on a cover of $U_i\cap U_j$
by open affines, $g_i|_{U_i\cap U_j}=g_j|_{U_i\cap U_j}$. We
can thus glue the morphism $g_i$ to obtain a morphism
$g:Z\rightarrow \tilde{X}$ such that $\psi\circ g=f$. The
morphism $g$ is evidently unique.
Now we can define the identification maps $\phi_{ij}$. Let
$\{U_i=\spec A_i\}$ be the open affine subsets of $X$. Let
$\{\tilde U_i=\spec \tilde A_i\}$ be the associated normalizations.
Let $\psi_i:\tilde U_i \rightarrow U_i$ be the morphism induced
by the inclusion $A_i \hookrightarrow \tilde A_i$. Let
$W_{ij}=\psi_i^{-1}(U_i\cap U_j)$. Then $W_{ij}$ is an open
subset of a normal scheme hence normal. $\psi_i:W_{ij}\rightarrow
U_i\cap U_j \subseteq U_j$ so there is a unique morphism
which we call $\phi_{ij}:W_{ij}\rightarrow \tilde U_j$ such
that $\psi_j|_{W_{ji}} \circ \phi_{ij} = \psi_i|_{W_{ij}}$.
By uniqueness we see that $\phi_{ij}\circ\phi_{ji}=id$ so
$\phi_{ij}=\phi_{ji}^{-1}$. Furthermore, for each $i,j,k$,
$\phi_{ij}(W_{ij}\cap W_{ik})
=\psi_j^{-1}(\psi_i(W_{ij}\cap W_{ik}))
=\psi_j^{-1}(\psi_i(W_{ij}))\cap \psi_j^{-1}(\psi_i(W_{ik}))
=W_{ji}\cap W_{jk}$. By uniqueness,
$\phi_{jk}\circ\phi_{ij}=\phi_{ik}$ on $W_{ij}\cap W_{ik}$.
So by the glueing lemma (Exercise 2.12) we may glue to obtain
a scheme $\tilde X$. We can also glue the morphisms $\psi_i$
to obtain a morphism $\psi:\tilde X\rightarrow X$.
Next, we must verify that the universal property holds in general.
Let $Z$ be an arbitrary normal integral scheme, and let $X$
and $\tilde{X}$ be as above and suppose $f:Z\rightarrow X$ is a morphism.
Cover $X$ be open affines $U_i$. Then for each morphism $f|_{f^{-1}(U_i)}$
we can apply the above proposition to find a morphism $g_i$ such
that $\psi\circ g_i=f|_{f^{-1}(U_i)}$. By uniqueness we can
glue these morphism to obtain the required morphism
$g:Z\rightarrow \tilde{X}$.
Now we check that $\tilde{X}$ is a normal integral scheme.
Note first that each $\tilde{U_i}$ is the spectrum of
an integrally closed domain and is hence a normal integral
scheme (since the localization of an integrally closed
domain is integrally closed).
Let $x\in\tilde{X}$. Then $x$ is contained in some $\tilde{U_i}$.
But the local ring of $x$ in $\tilde{X}$ is the same as the
local ring of $x$ in $\tilde{U_i}$ which is integrally closed.
This shows that $\tilde{X}$ is normal.
Since $X$ is irreducible,
every $U_i$ intersects every $U_j$. Thus every $\tilde{U_i}$
intersects every $\tilde{U_j}$ after glueing.
Since each $\tilde{U_j}$ is irreducible and they all overlap
this implies $\tilde{X}$ is irreducible. Indeed, if $\tilde{X}=A\cup B$
with $A$ and $B$ closed, then every $\tilde{U_i}$ is either completely
contained in $A$ or in $B$. If they are not all contained in one of
$A$ or $B$ then we can find an open subset $U$ contained in $A$
and an open subset $V$ contained in $B$ but not contained in $A$. Then
$V=(U\cap V)\cup(U^c\cap V)
=(A\cap V)\cup(U^c\cap V)$
which expresses $V$ as a union of two proper closed subsets of $V$.
$A\cap V$ is a proper subset of $V$ since $V$ is not contained in
$A$ and $U^c\cap V$ is a proper subset of $V$ since
$U\cap V\not=\emptyset$. This contradicts the fact that $V$ is
irreducible. Thus $\tilde{X}=A$ or $\tilde{X}=B$ whence $\tilde{X}$
is irreducible.
Now we check that the structure sheaf has no nilpotents.
Let $U$ be an open subset of $\tilde{X}$ and suppose
$f\in\so_{\tilde{X}}(U)$ is nilpotent. Then since $f$ is
nonzero, there is some point $x\in \tilde{X}$ so that the
stalk $f_x$ of $f$ at $x$ in the local ring $\so_{\tilde{X}}$ is
nonzero and nilpotent (use sheaf axiom (iii) and the definition of
$\so_{\tilde{X}}$.) Let $\tilde{U_i}$ be some $\tilde{U_i}$
which contains $x$. Then the local ring at $x$ is a localization
of the integral domain $\tilde{A_i}$ so it can't contain any
nilpotents. Thus the scheme $\tilde{X}$ is reduced.
Now we check that if $X$ is of finite type over a field $k$, then
the morphism $f:\tilde X\rightarrow X$ is a finite morphism.
The $U_i$ form an open cover of $X$ and $f^{-1}(U_i)=\spec \tilde A_i$
is affine for each $i$, so we just need to check that
$\tilde A_i$ is a finite module over $A_i$. This follows from
Theorem 3.9A of chapter I.
\begin{prob}[3.12]
Closed Subschemes of $\proj S$.
(a) Let $\phi:S\rightarrow T$ be a surjective homomorphism of graded
rings, preserving degrees. Show that the open set $U$ of (Ex. 2.14)
is equal to $\proj T$, and the morphism $f:\proj T\rightarrow\proj S$
is a closed immersion.
(b) If $I\subseteq S$ is a homogeneous ideal, take $T=S/I$ and let $Y$
be the closed subscheme of $X=\proj S$ defined as the image of the
closed immersion $\proj S/I\rightarrow X$. Show that different homogeneous
ideals can give rise to the same closed subscheme. For example,
let $d_0$ be an integer, and let $I'=\bigoplus_{d\geq d_0} I_d$. Show
that $I$ and $I'$ determine the same closed subscheme.
\end{prob}
\proof
(a) Since $\phi$ is graded and surjective, $\phi(S_{+})=T_{+}$ from which
it is immediate that $U=\proj T$. By the first isomorphism theorem
$T\cong S/\ker(\phi)$ so $f(\proj T)=f(\proj S/\ker(\phi))
=V(\ker(\phi))$ which is a closed subset of $\proj S$. (This is
just the fact that there is a one to one correspondence between
homogeneous ideals of $S/\ker(\phi)$ and homogeneous ideals of $S$
which contain $\ker(\phi)$.) The map on the stalk corresponding
to a point $x\in\proj T$ is the map
$S_{(\phi^{-1}(x))}\rightarrow T_{(x)}$
induced by $\phi$. This map is surjective since $\phi$ is surjective.
Thus the induced map on sheaves is surjective.
(b) Let $\phi:S/I'\rightarrow S/I$ be the natural projection homomorphism.
(This makes sense because $S/I$ is a quotient of $S/I'$. Indeed,
$S/I=(S/I')/\bigoplus_{0\leq d